ExamVeda
Login
Home

The value of the definite integral int 1^ text e sqrt...

This question belongs to Engineering Maths Calculus
Calculus
?

The value of the definite integral $$\int_1^{\text{e}} {\sqrt {\text{x}} } \ln \left( {\text{x}} \right){\text{dx}}$$    is

Answer & Solution
Correct Answer: Option C
Let's discuss the solution. Join the discussion
Examveda
Question posted by Examveda
Community

Join the Discussion

No comments yet Be the first to discuss this question.
Keep Learning

Related Questions on Calculus

What is the value of \[\mathop {\mathop {\lim }\limits_{{\text{x}} \to 0} }\limits_{{\text{y}} \to 0} \frac{{{\text{xy}}}}{{{{\text{x}}^2} + {{\text{y}}^2}}}?\]
A. 1 B. -1 C. 0 D. Limit does not exit
The Taylor series expansion of 3 sinx + 2 cosx is . . . . . . . .
A. 2 + 3x - x2 - \[\frac{{{{\text{x}}^3}}}{2}\] + ... B. 2 - 3x + x2 - \[\frac{{{{\text{x}}^3}}}{2}\] + ... C. 2 + 3x + x2 + \[\frac{{{{\text{x}}^3}}}{2}\] + ... D. 2 - 3x - x2 + \[\frac{{{{\text{x}}^3}}}{2}\] + ...
\[\mathop {\lim }\limits_{{\text{x}} \to \infty } \sqrt {{{\text{x}}^2} + {\text{x}} - 1} - {\text{x is}}\]
A. 0 B. \[\infty \] C. \[\frac{1}{2}\] D. \[ - \infty \]
The Taylor series expansion of \[\frac{{\sin {\text{x}}}}{{{\text{x}} - \pi }}\]  at \[{\text{x}} = \pi \]  is given by
A. \[1 + \frac{{{{\left( {{\text{x}} - \pi } \right)}^2}}}{{3!}} + ...\] B. \[ - 1 - \frac{{{{\left( {{\text{x}} - \pi } \right)}^2}}}{{3!}} + ...\] C. \[1 - \frac{{{{\left( {{\text{x}} - \pi } \right)}^2}}}{{3!}} + ...\] D. \[ - 1 + \frac{{{{\left( {{\text{x}} - \pi } \right)}^2}}}{{3!}} + ...\]
More Related Questions on Calculus